One other way you can prove all diagram chasing results (that I know of) is to first prove that there is a spectral sequence associated with a double complex (actually, there are two!). This involves some diagram chasing, but you only do it once. I guess you could also argue that you still have to chase the spectral sequences.
As an example, here's how you'd prove that a short exact sequence of chain complexes induces a long exact sequence in homology. We start with the double complex with exact rows
↑ ↑ ↑
0 → Ai+1 → Bi+1 → Ci+1 → 0
↑ ↑ ↑
0 → Ai → Bi → Ci → 0
↑ ↑ ↑
(I'm assuming the columns are bounded below, but it doesn't matter since this result is "local": you can truncate above and below the point you're interested in and then prove the result)
We can regard this as the $E_0$ page of two different spectral sequences, depending on whether we decide to start with the vertical or horizontal arrows as our differential. Both spectral sequences have to abut to the homology of the total complex associated to this double complex.
First, use the horizontal arrows. Since the rows are exact, when we take homology, we get zero everywhere, so the $E_1$ page is identically zero. Nothing interesting is going to happen now: we've gotten to $E_\infty$. So the homology of the total complex is zero.
Now what if we use the vertical arrows? We get the $E_1$ page
0 → Hi+1(A) → Hi+1(B) → Hi+1(C) → 0
0 → Hi(A) → Hi(B) → Hi(C) → 0
We'd like to prove that the rows are exact in the middle and that kernel of $H^{i+1}(A)\to H^{i+1}(B)$ is isomorphic to the cokernel of $H^i(B)\to H^i(C)$. Let's flip to the $E_2$ page and see what happens. Let $K^i=\ker(H^{i+1}(A)\to H^{i+1}(B))$, $M^i=\mathrm{cok}(H^i(B)\to H^i(C))$, and let $L^i$ be the homology of the above row at $H^i(B)$. We get the $E_2$ page
0 Ki+1 Li+1 Mi+1 0
↘ ↘ ↘
↘ ↘ ↘
0 Ki Li Mi 0
(pardon my ascii art: those are meant to be arrows going two spots to the right and one spot down)
Note that on the $E_{\ge 3}$ pages, the differentials will be too long to connect anything in these three columns to anything other than zero. Since we must abut to zero, this $E_2$ page is the "last chance" for the homology to vanish. It follows that the sequences $0\to L^i\to 0$ and $0\to K^{i+1}\to M^i\to 0$ must be exact, so $L^i=0$ and $K^{i+1}\cong M^i$, which is exactly what we wanted to show.
Note that I was able to completely ignore the question of what the differentials on higher pages were; I just had to know that they exist.
Diagram-chasing results almost always assume that the rows are exact and then make assertions about the homology groups you get from the columns, so you can always run one spectral sequence and immediately get that the homology of the total complex is zero, then run the other spectral sequence and note when the spectral sequence has it's "last chance" to cancel all homology groups.
Very often one has the feeling that set-theoretic issues are somewhat cheatable, and people feel like they have eluded foundations when they manage to cheat them. Even worse, some claim that foundations are irrelevant because each time they dare to be relevant, they can be cheated. What these people haven't understood is that the best foundation is the one that allows the most cheating (without falling apart).
In the relationship between foundation and practice, though, what matters the most is the phenomenology of every-day mathematics. In order to make this statement clear, let me state the uncheatable lemma. In the later discussion, we will see the repercussion of this lemma.
Lemma (The uncheatable).
A locally small, large-cocomplete category is a poset.
The lemma shows that no matter how fat are the sets where you enrich your category, there is no chance that the category is absolutely cocomplete.
Example. In the category of sets, the large coproduct of all sets is not a set. If you enlarge the universe in such a way that it is, then some other (even larger) coproduct will not exist. This is inescapable and always boils down to the Russel Paradox.
Remark. Notice that obvious analogs of this lemma are true also for categories based on Grothendieck Universes (as opposed to sets and classes). One can't escape the truth by changing its presentation.
Excursus. Very recently Thomas Forster, Adam Lewicki, Alice Vidrine have tried to reboot category theory in Stratified Set Theory in their paper Category Theory with Stratified Set Theory (arXiv: https://arxiv.org/abs/1911.04704). One could consider this as a kind of solution to the uncheatable lemma. But it's hard to tell whether it is a true solution or a more or less equivalent linguistic reformulation. This theory is at its early stages.
At this point one could say that I haven't shown any concrete problem, we all know that the class of all sets is not a set, and it appears as a piece of quite harmless news to us.
In the rest of the discussion, I will try to show that the uncheatable lemma has consequences in the daily use of category theory. Categories will be assumed to be locally small with respect to some category of sets. Let me recall a standard result from the theory of Kan extensions.
Lemma (Kan). Let $\mathsf{B} \stackrel{f}{\leftarrow} \mathsf{A} \stackrel{g}{\to} \mathsf{C}$ be a span where $\mathsf{A}$ is small and $\mathsf{C}$ is (small) cocomplete. The the left Kan extension $\mathsf{lan}_f g$ exists.
Kan extensions are a useful tool in everyday practice, with applications in many different topics of category theory. In this lemma (which is one of the most used in this topic) the set-theoretic issue is far from being hidden: $\mathsf{A}$ needs to be small (with respect to the size of $\mathsf{C})$! There is no chance that the lemma is true when $\mathsf{A}$ is a large category. Indeed since colimits can be computed via Kan extensions, the lemma would imply that every (small) cocomplete category is large cocomplete, which is not allowed by the uncheatable. Also, there is no chance to solve the problem by saying: well, let's just consider $\mathsf{C}$ to be large-cocomplete, again because of the the uncheatable.
This problem is hard to avoid because the size of the categories of our interest is as a fact always larger than the size of their inhabitants (this just means that most of the time Ob$\mathsf{C}$ is a proper class, as big as the size of the enrichment).
Notice that the Kan extension problem recovers the Adjoint functor theorem one, because adjoints are computed via Kan extensions of identities of large categories, $$\mathsf{R} = \mathsf{lan}_\mathsf{L}(1) \qquad \mathsf{L} = \mathsf{ran}_\mathsf{R}(1) .$$ Indeed, in that case, the solution set condition is precisely what is needed in order to cut down the size of some colimits that otherwise would be too large to compute, as can be synthesized by the sharp version of the Kan lemma.
Sharp Kan lemma. Let $\mathsf{B} \stackrel{f}{\leftarrow} \mathsf{A} \stackrel{g}{\to} \mathsf{C}$ be a span where $\mathsf{B}(f-,b)$ is a is small presheaf for every $b \in \mathsf{B}$ and $\mathsf{C}$ is (small) cocomplete. Then the left Kan extension $\mathsf{lan}_f g$ exists.
Indeed this lemma allows $\mathsf{A}$ to be large, but we must pay a tribute to its presheaf category: $f$ needs to be somehow locally small (with respect to the size of $\mathsf{C}$).
Kan lemma Fortissimo. Let $ \mathsf{A} \stackrel{f}{\to} \mathsf{B} $ be a functor. The following are equivalent:
- for every $g :\mathsf{A} \to \mathsf{C}$ where $\mathsf{C}$ is a small-cocomplete category, $\mathsf{lan}_f g$ exists.
- $\mathsf{lan}_f y$ exists, where $y$ is the Yoneda embedding in the category of small presheaves $y: \mathsf{A} \to \mathcal{P}(\mathsf{A})$.
- $\mathsf{B}(f-,b)$ is a is small presheaf for every $b \in \mathsf{B}$.
Even unconsciously, the previous discussion is one of the reasons of the popularity of locally presentable categories. Indeed, having a dense generator is a good compromise between generality and tameness. As an evidence of this, in the context of accessible categories the sharp Kan lemma can be simplified.
Tame Kan lemma. Let $\mathsf{B} \stackrel{f}{\leftarrow} \mathsf{A} \stackrel{g}{\to} \mathsf{C}$ be a span of accessible categories, where $f$ is an accessible functor and $\mathsf{C}$ is (small) cocomplete. Then the left Kan extension $\mathsf{lan}_f g$ exists (and is accessible).
Warning. The proof of the previous lemma is based on the density (as opposed to codensity) of $\lambda$-presentable objects in an accessible category. Thus the lemma is not valid for the right Kan extension.
References for Sharp. I am not aware of a reference for this result. It can follow from a careful analysis of Prop. A.7 in my paper Codensity: Isbell duality, pro-objects, compactness and accessibility. The structure of the proof remains the same, presheaves must be replaced by small presheaves.
References for Tame. This is an exercise, it can follow directly from the sharp Kan lemma, but it's enough to properly combine the usual Kan lemma, Prop A.1&2 of the above-mentioned paper, and the fact that accessible functors have arity.
This answer is connected to this other.
Best Answer
Here's an attempt, assuming the target category is finite dimensional vector spaces and the diagram is finite acyclic.
1) Fill in the diagram so it's triangulated (that is, add the compositions of all arrows). Add kernels and cokernels to every arrow in the diagram. Add the natural arrows between these new objects (e.g. the kernel of $f$ maps into the kernel of $g\circ f$). Iterate this process until it terminates (which it does, as there are finitely many arrows, and thus only finitely many subquotients of objects they induce).
2) In this new diagram, tabulate all exact paths. (This step seems to me to have high time and space complexity, unfortunately.) By exact path, I mean an infinite long exact sequence, for which all but finitely many objects are zero.
3) Given an exact path $$0\to A_1\to \cdots \to A_i\to \cdots \to A_n\to 0$$ write the equation $$\sum_{i=1}^n (-1)^i \dim~ A_i=0.$$
4) Solve the resulting linear system for the $\dim~ A_i$'s.
Since we've added kernels and cokernels, solving the system tells us about injectivity and surjectivity of all the arrows. Furthermore, as far as I can tell any (non-negative) solution to this linear system is realizable as a diagram, so this algorithm gives us any information we could get by diagram chasing.
Greg has given a good summary of this method in his answer below--he takes me to task about my assertion that any non-negative integer solution to this linear system is realizable, so I will sketch an argument (again requiring the diagram to be finite acyclic). We also assume the diagram is "complete" in the sense that step 1) has been completed.
Partially order the objects in the diagram by saying $A\leq B$ iff $A$ is a subquotient of $B$; inductively, this is the same as saying that $B\leq B$ and $A\leq B$ if there exists $C\leq B$ such that $A$ injects into $C$ or $C$ surjects onto $A$ in this diagram. We identify isomorphic objects; this is a poset by acyclicity. We proceed by induction on the number of objects in the diagram. The one-object diagram is easy, so we do the induction step.
By finiteness the diagram contains a longest chain; choose such a chain and consider its minimal element $X$. All maps into or out of $X$ are surjections or injections, respectively, so split $X$ off of every object that maps onto it or that it maps into as a direct summand. The diagram with $X$ removed gives the induction step.